Using an appropriate failure theory, find the factor of safety in each case. State the name of the theory that you are using the theory is max stress theory.
<h3>Wat is the max stress theory?</h3>
The most shear strain concept states that the failure or yielding of a ductile fabric will arise whilst the most shear strain of the fabric equals or exceeds the shear strain fee at yield factor withinside the uniaxial tensile test.”
Stress states at various critical locations are f= 2.662.
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Answer:
There is 0.466 KW required to operate this air-conditioning system
Explanation:
<u>Step 1:</u> Data given
Heat transfer rate of the house = Ql = 755 kJ/min
House temperature = Th = 24°C = 24 +273 = 297 Kelvin
Outdoor temperature = To = 35 °C = 35 + 273 = 308 Kelvin
<u>Step 2: </u> Calculate the coefficient of performance o reversed carnot air-conditioner working between the specified temperature limits.
COPr,c = 1 / ((To/Th) - 1)
COPr,c = 1 /(( 308/297) - 1)
COPr,c = 1/ 0.037
COPr,c = 27
<u>Step 3:</u> The power input cna be given as followed:
Wnet,in = Ql / COPr,max
Wnet, in = 755 / 27
Wnet,in = 27.963 kJ/min
Win = 27.963 * 1 KW/60kJ/min = 0.466 KW
There is 0.466 KW required to operate this air-conditioning system
Complete Question
The complete question is shown on the first uploaded image
Answer:
a) The required additional minterms for f so that f has eight primary implicants with two literals and no other prime implicant are 
and 
b) The essential prime implicant are 
and 
c) The minimum sum-of-product expression for f are
Explanation:
The explanation is shown on the second third and fourth image
Answer:
Headlights are required to be used 1/2 hour after sunset to 1/2 hour before sunrise, when windshield wipers are being used, when visibility is less than 1000 feet, or when there is insufficient light or adverse weather.
Explanation:
hope this helps
Answer:
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
Explanation:
We are given;
T∞ = 70°C.
Inner radii pipe; r1 = 6cm = 0.06 m
Outer radii of pipe;r2 = 6.5cm=0.065 m
Electrical heat power; Q'_s = 300 W
Since power is 300 W per metre length, then; L = 1 m
Now, to the heat flux at the surface of the wire is given by the formula;
q'_s = Q'_s/A
Where A is area = 2πrL
We'll use r2 = 0.065 m
A = 2π(0.065) × 1 = 0.13π
Thus;
q'_s = 300/0.13π
q'_s = 734.56 W/m²
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
